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Question Number 143475 by Ghaniy last updated on 14/Jun/21

 evaluate;  ((((√7))^(log64) −(3)^(log_(24) 8) )/((log _2 8−log _(1/4) 64)((1/(log _4 ((1/(64))))))))

$$\:{evaluate}; \\ $$$$\frac{\left(\sqrt{\mathrm{7}}\right)^{\mathrm{log64}} −\left(\mathrm{3}\right)^{\mathrm{log}_{\mathrm{24}} \mathrm{8}} }{\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{8}−\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{64}\right)\left(\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{64}}\right)}\right)} \\ $$

Commented by amin96 last updated on 14/Jun/21

super

$${super} \\ $$

Answered by Ar Brandon last updated on 15/Jun/21

ℵ=((((√7))^(log_(49) 64) −(3)^(log_(24) 8) )/((log_2 8−log_(1/4) 64)((1/(log_4 ((1/(64))))))))     =((7^((1/2)∙(1/2)log_7 64) −3^(log_(24) 8) )/((3+3)(−(1/3))))=−((64^(1/4) −3^(log_(24) 8) )/2)      =(1/2)∙3^(log_(24) 8) −(√2)=...

$$\aleph=\frac{\left(\sqrt{\mathrm{7}}\right)^{\mathrm{log}_{\mathrm{49}} \mathrm{64}} −\left(\mathrm{3}\right)^{\mathrm{log}_{\mathrm{24}} \mathrm{8}} }{\left(\mathrm{log}_{\mathrm{2}} \mathrm{8}−\mathrm{log}_{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{64}\right)\left(\frac{\mathrm{1}}{\mathrm{log}_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{64}}\right)}\right)} \\ $$$$\:\:\:=\frac{\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{7}} \mathrm{64}} −\mathrm{3}^{\mathrm{log}_{\mathrm{24}} \mathrm{8}} }{\left(\mathrm{3}+\mathrm{3}\right)\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)}=−\frac{\mathrm{64}^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{3}^{\mathrm{log}_{\mathrm{24}} \mathrm{8}} }{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{3}^{\mathrm{log}_{\mathrm{24}} \mathrm{8}} −\sqrt{\mathrm{2}}=... \\ $$

Commented by Ghaniy last updated on 15/Jun/21

thanks

$${thanks} \\ $$

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